Saturday, August 12, 2017

Practising coding problems are very important to do well in any programming interview. You should at your best on data-structures like an array, linked list, and string to clear any programming interview and believe me, you can not do this in one day or one week. It's rather a long process of learning through coding, and that's where these small coding problems help. Today, we are going to look at another interesting programming question from the array; write a program to find all pairs of integers whose sum is equal to a given number. For example if input integer array is {2, 6, 3, 9, 11} and given sum is 9, output should be {6,3}. Sounds simple? maybe, but this exact question has appeared in a technical interview at Amazon, Microsoft, Facebook and couple of another fortune five tech companies in past. Many of you might already heard about this question and some of you may already know the solution to this problem as well, but it's not enough to know just the answer. In a programming interview, many things matter apart from correct solution. For example, first thing Interviewer look is whether a candidate can ask right questions or not. So before jumping straight to coding, spare a second or two to think about the problem and clear any doubt you may have. For example, you can ask following questions based upon problem statement given above :

Does array contains only positive or negative numbers?

What if the same pair repeats twice, should we print it every time?

Is reverse of pair is acceptable e.g. can we print both (4,1) and (1,4) if given sum is 5.

Do we need to print only distinct pair? does (3, 3) is a valid pair forgiven sum of 6?

How big the array is?

Many programmers afraid to ask questions instead they like to assume about it, but during coding interview IMHO it's always better to ask questions. First, it shows that you have not mugged the answer and second it demonstrates that you have the ability to think through a problem, which is a very important quality of any professional programmer. Now let's go back to question, for simplicity we can assume that we just need to print a pair of integers once or twice depending upon their occurrence, but pair has to be distinct, (2,2) or (3, 3) is not valid pair.

3 Solution to Find Pair Of Integers in Array whose Sum is Given Number

The first solution which comes in my mind is our friend brute-force, naive but genuine. You take one number from array and then loop through array and output pairs which is equal to given sum. You do this for all numbers in first array, as shown in following Java program :

This solution is correct but it's time complexity is very hight, O(n^2), which means Interviewer will surely ask you to improve your answer and come up with solution whose complexity is either O(1), O(n) or O(nLog(n)). So let's dig deeper to improve this answer. In order to find two numbers in an array whose sum equals a given value, we probably don't need compare each number with other. What we can do here is to store all numbers in a hashtable and just check if it contains second value in a pair. For example, if given sum is 4 and one number in pair is 3, then other must be 1 or -7. Do you remember the first question we asked, if array only contains positive numbers then we don't need to check for negative values in Map. How is this solution better than previous one? It would require less comparisons. Only N to iterate through array and insert values in a Set because add() and contains() both O(1) operation in hash table. So total complexity of solution would be O(N). Here is a Java program which find the pair of values in the array whose sum is equal to k using Hashtable or Set. In this program we have also written a utility method to generate random numbers in a given range in Java. You can use this method for testing with random inputs. By the way, random numbers are only good for demonstration, don't use them in your unit test. One more good thing you can learn from printPairsUsingSet() method is pre validation, checking if inputs are valid to proceed further.

One more thing, here we are using HashSet but since HashSet in Java internally uses HashMap, it would not make any difference if use either of those data structure.By the this solution has few constraints, first it would need additional space of order O(n) to store numbers in Hashtable or Set, so you need additional space which could be problem if array is very large (remember the question we asked before writing solution). For a large array, you need a solution which doesn't require additional space, also known as in-place solution. If interviewer will ask you how do you find if two values in an array sum to a given value without any additional space, first solution will also not work because it's complexity is too high and it would too long to sort a large array. A solution with complexity e.g. O(n), O(logN) or O(NLongN) should work though. A more efficient in-place solution would be to sort the array and use two pointers to scan through array from both direction i.e. beginning and end. If sum of both the values are equal to given number then we output the pair and advance them. If the sum of two numbers is less than k then we increase the left pointer, else if the sum is greater than k we decrement the right pointer, until both pointers meet at some part of the array. The complexity of this solution would be O(NlogN) due to sorting. Remember to use a in-place sorting algorithm like quicksort to sort the array as we don't have additional space. Thankfully, Arrays.sort() method uses a two pivot quicksort algorithm to sort array of primitives.

That' all on this array based interview question to find all pairs in an array of integers whose sum is equal to a given integer. We have seen three ways to solve this problem starting from simplest brute-force solution to acceptable O(N) with additional space and O(NLogN) in-place. If anyone like to do some more practice, I would suggest to write JUnit test cases for this problem, given set of constraints that only unique pair needs to be printed even if array contains duplicated and find bugs on these solution. Alternatively, you can also try to solve it's cousin question, given an array of integers check whether there are 3 numbers that sum up to 0 or given number. Remember more fun is in journey than reaching the destination :)

Exercises :
1) Write JUnit tests for this problem and check if each of this solution passes those tests.
2) Come up with a better solution in terms of time and space complexity?
3) Find boundary conditions on which these solution breaks.

19 comments
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In a solution involving HashSet where if a target number is not in the set it is added to it you are checking if a set contains a number with contains but if I read correctly the .add method returns false if the object you are adding is already in the set so this piece of code:

The condition i<=(sum/2) is required to prevent double repetition of pairs (2 & 5) and (5 & 2). The condition i!=(sum-i) is required to prevent same number in paring. (For example, if your sum is 8, then if you don't use this condition, you will also get pair (4,4) ).

A perfect number is one that is the sum of its factors, excluding itself. The 1st perfect number is 6 because 6 = 1 + 2 + 3. The 2nd perfect number is 28 which equals 1 + 2 + 4 + 7 + 14. The third is 496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248. In each case, the number is the sum of all its factors excluding itself.

Write a method named henry that takes two integer arguments, i and j and returns the sum of the ith and jth perfect numbers. So for example, henry (1, 3) should return 502 because 6 is the 1st perfect number and 496 is the 3rd perfect number and 6 + 496 = 502.

The function signature is int henry (int i, int j)

You do not have to worry about integer overflow, i.e., you may assume that each sum that you have to compute can be represented as a 31 bit integer. Hint: use modulo arithmetic to determine if one number is a factor of another.

In the section when you explain the Hash method, the example is wrong: " For example, if given sum is 4 and one number in pair is 3, then other must be 1 or -7."The other can't be -7, otherwise the sum is -4 not 4.